Question

Find the modulus and amplitude of $-\sqrt{3}-i$
$\begin{align}
  & a)|z|=2;amp(z)=-\dfrac{5\pi }{6} \\
 & b)|z|=4;amp(z)=\dfrac{5\pi }{6} \\
 & c)|z|=4;amp(z)=-\dfrac{\pi }{6} \\
 & \text{d) None of these} \\
\end{align}$

Answer 1

Hint: For a complex function of the form a + ib, we have the modulus of the complex number is \[\sqrt{{{a}^{2}}+{{b}^{2}}}\] . Now to find the amplitude first we will find the value of ${{\tan }^{-1}}\left( \dfrac{b}{a} \right)$ .

Complete step by step answer:
Now let us consider the given complex number.
The given complex number is $-\sqrt{3}-i$
Now we can also write this complex number as \[\left( -\sqrt{3} \right)+\left( -1 \right)i\]
Now comparing this equation with the general form of complex numbers which is a + bi, we get
a = $-\sqrt{3}$ and b = $-1$ .
Let us first calculate the modulus of the complex number.
We know that modulus of the complex number of the form a + ib is \[\sqrt{{{a}^{2}}+{{b}^{2}}}\].
Hence we get the modulus of given complex number is \[\sqrt{{{\left( -\sqrt{3} \right)}^{2}}+{{\left( -1 \right)}^{2}}}\]
\[\sqrt{{{\left( -\sqrt{3} \right)}^{2}}+{{\left( -1 \right)}^{2}}}=\sqrt{3+1}=\sqrt{4}=2\]
Hence the modulus of a given complex number is 2.
Now let us find the argument of the given complex number.
To find the argument of any complex number of the form a + ib we first need to find ${{\tan }^{-1}}\left( \dfrac{b}{a} \right)$
Hence we get ${{\tan }^{-1}}\left( \dfrac{-1}{-\sqrt{3}} \right)=\dfrac{\pi }{6}$ .
Now since we have $0<\dfrac{\pi }{6}<\dfrac{\pi }{2}$ .
Now since the point is in third quadrant the angle will be given by
$\dfrac{\pi }{6}-\pi =\dfrac{\pi -6\pi }{6}=-\dfrac{5\pi }{6}$ .
Hence the modulus of complex numbers is 2 and the amplitude is $-\dfrac{5\pi }{6}$ .

So, the correct answer is “Option a”.

Note: Now note that the amplitude is not always equal to ${{\tan }^{-1}}\left( \dfrac{b}{a} \right)$ . It depends on which quadrant, the complex number lies. For a complex number in the first quadrant the amplitude is ${{\tan }^{-1}}\left( \dfrac{b}{a} \right)$ . If the complex number lies in the second quadrant then the amplitude is given by $\pi -{{\tan }^{-1}}\left( \dfrac{b}{a} \right)$ . Now is the point lies in third quadrant then the argument is ${{\tan }^{-1}}\left( \dfrac{b}{a} \right)-\pi $ and for fourth quadrant the amplitude is $-{{\tan }^{-1}}\left( \dfrac{b}{a} \right)$ .